用无穷级数评估sinx [英] Evaluation of sinx using infinite series

问题描述

使用无限系列评估sinx



我尝试过：

evaluation of sinx using infinite series

What I have tried:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()
{
    int terms;
    int count=0;
    int sum = 0;int x;int n;
    printf("enter number of terms");
    scanf("%d",&terms);
    printf("enter value of x");
    scanf("%d",&x);
    while(count <= terms){

        n = 2*count+1;

        sum = sum + ((pow(-1,count))*(pow(x,n)/(fact(n))));
        count++;
    }
    printf("%d",sum);
    }
    int fact(int f){
    int number = 1;
    int product = 1;
    while(number<=f){

         product = number*product;
         number++;

    }
    return product;
}

推荐答案

使用无限级数计算函数需要很长时间。在此之前，问问自己：



1.函数是否定期？如果是这样，将参数减少到一个句点。

2.是奇数还是偶数（f（x）== -f（-x）或f（x）== f（-x ））？如果是这样，写奇数函数，例如x * F（x ^ 2），偶数函数，例如G（x ^ 2）。

3.是否有任何其他属性可以提供帮助的功能吗？



完成后，您会发现只需要有限数量的元素即可获得所需的精确度。



祝你好运！

Using an infinite series to calculate a function will take a long time. Before doing so, ask yourself:

1. Is the function periodic? If so, reduce the argument to a single period.
2. Is it odd or even (f(x) == -f(-x) or f(x) == f(-x) )? If so, write the odd function as for example x*F(x^2), and the even function as for example G(x^2).
3. Are there any other attributes of the function that can help?

Once you have done that, you will find that only a finite number of elements are required to get the accuracy that you want.

Good luck!

根据建议，使用 double 而不是 INT 。享受：

As suggested, use doubles instead of int. Enjoy:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

double fact(unsigned n)
{
  double result =1.0;
  unsigned int i;
  for (i=2; i<=n; ++i )
    result *= i;
  return result;
}

int main()
{
  unsigned terms;
  unsigned count=0;
  double result = 0.0;
  double angle;

  printf("enter number of terms ");
  scanf("%d",&terms);
  printf("enter value of x ");
  scanf("%lf",&angle);
  while(count <= terms)
  {
      double sign = count & 1 ? -1.0 : 1.0;
      unsigned n = 2*count+1;

      result += sign*(pow(angle,(double)n)/fact(n));
      count++;
  }
  double br = sin(angle);
  printf("approximation %g, builtin function result %g, difference  %5.2f%%", result, br, fabs(result-br)/br*100.0);
}

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